How many liters of a .88 M solution can be made with 25.5 grams of lithium fluoride (#LiF#)?
For this question we are given the Molarity 0.88M
We are told the solute is a 25.2 gram sample of LiF, Lithium Fluoride
We can convert the mass of LiF to moles by dividing by the molar mass of LiF
Li = 6.94
F = 19.0
LiF = 25.94 g/mole
Now we can take the the molarity and the moles and calculate the Liters of solution